A university surveyed recent graduates of the English department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000. The population standard deviation was $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English department

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.

Number of sample, n = 400

Mean, u = $25,000

Standard deviation, s = $2,500

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z * standard deviation/√n

It becomes

25000 ± 1.96 * 2500/√400

= 25000 ± 1.96 * 125

= 25000 ± 245

The lower end of the confidence interval is 25000 - 245 = 24755

The upper end of the confidence interval is 25000 + 245 = 25245

Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245